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## Homework Statement

Find the the acceleration,

*a[/i, and the Tension,*

Other relevant information:

Mass 1 is a 5 or 2 kg block on a 30 degree incline plane (based so that Fy is Fsin(30) ) connected to a massless rope that holds Mass 2, a 5 kg block, over a pulley.

Ff = (µk)(Fn)

w = mg (acceleration due to gravity is assumed to be 10 m/s^2 for easy calculation according to the professor)

Fnet = ma

I solved for the weight of mass 1 as being:

m1 = 5 kg: W = 50 N * sin(30) = 25 N

m2 = 2 kg: W = 20 N * sin(30) = 10 N

And found force of friction:

Ff = 0.1 * 25 N = 2.5 N

The weight of block m2:

w = 10 m/s^2 * 5 kg = 50 N

I tried creating a FBD for block m1, where the following forces were applied:

W = 25 N

Fn = 25 N

Ff = 2.5 N

F = Wm2 (Wm2 = 50 N)

And a FBD for block m2, where:

W = 50 N

Except, I'm confused where to go from here now. I know I have to calculate tension still. I'm thinking that the total force being applied on m1 according to Newton's 2nd Law is:

Fnet = ma - Ff = 50 N - 2.5 N = 47.5 N

And acceleration is: a = 47.5 N / 5 kg = 9.5 m/s^2

2nd attempt:

I know Fnet = ma - T. and Fnet is not 0 N, it's 50 N. Since the rope is massless the tension is 2T = 100 N.

I therefore arrive at the following answers of:

a = 9.5 m/s^2

T = 100 N

Does this look right?

*T*, in the system shown below if (a) m1 = 5 kg and (b) m1 = 2 kg. assume the coefficient of kinetic fricton on the incline plane is µk = 0.1.Other relevant information:

Mass 1 is a 5 or 2 kg block on a 30 degree incline plane (based so that Fy is Fsin(30) ) connected to a massless rope that holds Mass 2, a 5 kg block, over a pulley.

## Homework Equations

Ff = (µk)(Fn)

w = mg (acceleration due to gravity is assumed to be 10 m/s^2 for easy calculation according to the professor)

Fnet = ma

## The Attempt at a Solution

I solved for the weight of mass 1 as being:

m1 = 5 kg: W = 50 N * sin(30) = 25 N

m2 = 2 kg: W = 20 N * sin(30) = 10 N

And found force of friction:

Ff = 0.1 * 25 N = 2.5 N

The weight of block m2:

w = 10 m/s^2 * 5 kg = 50 N

I tried creating a FBD for block m1, where the following forces were applied:

W = 25 N

Fn = 25 N

Ff = 2.5 N

F = Wm2 (Wm2 = 50 N)

And a FBD for block m2, where:

W = 50 N

Except, I'm confused where to go from here now. I know I have to calculate tension still. I'm thinking that the total force being applied on m1 according to Newton's 2nd Law is:

Fnet = ma - Ff = 50 N - 2.5 N = 47.5 N

And acceleration is: a = 47.5 N / 5 kg = 9.5 m/s^2

2nd attempt:

I know Fnet = ma - T. and Fnet is not 0 N, it's 50 N. Since the rope is massless the tension is 2T = 100 N.

I therefore arrive at the following answers of:

a = 9.5 m/s^2

T = 100 N

Does this look right?

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